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THE MOLETHE MOLEA guide for AS level studentsA guide for AS level students
KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING20152015
SPECIFICATIONSSPECIFICATIONS
OBJECTIVE
• The foundations of chemistry• Formulae and equations• The mole• Using moles –reacting masses• Calculating formulae using moles• Measuring concentration – molarity• Measuring concentration – other units• More ways to calculate equations
INTRODUCTION
This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at...
www.knockhardy.org.uk/sci.htm
Navigation is achieved by using the left and right arrow keys on the keyboard
KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING
THE MOLETHE MOLE
CONTENTS
• What is a mole and why do we use it?
• Calculating the number of moles of a single substance
• Reacting mass calculations
• Solutions and moles
• Standard solutions
• Volumetric calculations
• Molar volume calculations
THE MOLETHE MOLE
Before you start it would be helpful to…
• know how to balance simple equations
• know how to re-arrange mathematical formulae
THE MOLETHE MOLE
DON’T BE LEFT IN THE
DARK!
WHAT IS A MOLE ?
it is the standard unit of amount of a substance - it is just a number, a very big numberit is a way of saying a number in words, just like...
DOZEN for 12 SCORE for 20 GROSS for 144
THE MOLETHE MOLE
HOW BIG IS IT ?
602200000000000000000000 (Approximately)... THAT’S BIG !!!
It is a lot easier to write it as... 6.022 x 1023
THE MOLETHE MOLE
WHAT IS A MOLE ?
it is the standard unit of amount of a substance - it is just a number, a very big numberit is a way of saying a number in words, just like...
DOZEN for 12 SCORE for 20 GROSS for 144
HOW BIG IS IT ?
602200000000000000000000 (Approximately)... THAT’S BIG !!!
It is a lot easier to write it as... 6.022 x 1023
It is also known as... AVOGADRO’S NUMBER
It doesn’t matter what the number is as long as everybody sticks to the same value !
THE MOLETHE MOLE
WHAT IS A MOLE ?
it is the standard unit of amount of a substance - it is just a number, a very big numberit is a way of saying a number in words, just like...
DOZEN for 12 SCORE for 20 GROSS for 144
WHY USE IT ?
Atoms and molecules don’t weigh much so it is easier to count large numbers of them. In fact it is easier to weigh substances.
Using moles tells you... how many particles you get in a certain mass the mass of a certain number of particles
DO I NEED TO KNOW ANYTHING ELSE ?
Yes, it would help if you can balance equations
AND Keep trying, you will get the idea ... EVENTUALLY!
THE MOLETHE MOLE
WHAT IS IT? The standard unit of amount of a substance - just as the standard unit of length is a METREIt is just a number, a very big numberIt is also a way of saying a number in wordslike DOZEN for 12
GROSS for 144
THE MOLE – AN OVERVIEWTHE MOLE – AN OVERVIEW
WHAT IS IT? The standard unit of amount of a substance - just as the standard unit of length is a METREIt is just a number, a very big numberIt is also a way of saying a number in wordslike DOZEN for 12
GROSS for 144
HOW BIG IS IT ? 602200000000000000000000 (approx) - THAT’S BIG !!!It is a lot easier to write it as 6.022 x 1023
And anyway it doesn’t matter what the number isas long as everybody sticks to the same value !
THE MOLE – AN OVERVIEWTHE MOLE – AN OVERVIEW
WHAT IS IT? The standard unit of amount of a substance - just as the standard unit of length is a METREIt is just a number, a very big numberIt is also a way of saying a number in wordslike DOZEN for 12
GROSS for 144
HOW BIG IS IT ? 602200000000000000000000 (approx) - THAT’S BIG !!!It is a lot easier to write it as 6.022 x 1023
And anyway it doesn’t matter what the number isas long as everybody sticks to the same value !
WHY USE IT ? Atoms and molecules don’t weigh much so it iseasier to count large numbers of them.In fact it is easier to weigh substances.
Using moles tells you :- how many particles you get in a certain massthe mass of a certain number of particles
THE MOLE – AN OVERVIEWTHE MOLE – AN OVERVIEW
CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE
moles = mass / molar mass mass = moles x molar mass
molar mass = mass / moles
UNITS
mass g or kgmolar mass g mol-1 or kg mol-1
THE MOLETHE MOLE
MOLES = MASS MOLAR MASS MOLES = MASS MOLAR MASS
MASS
MOLES x MOLAR MASS
COVER UP THE VALUE YOU WANT AND THE METHOD
OF CALCULATION IS REVEALED
1. Calculate the number of moles of oxygen molecules in 4g
MOLES OF A SINGLE SUBSTANCEMOLES OF A SINGLE SUBSTANCE
MASS
MOLES x MOLAR MASS
1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
MOLES OF A SINGLE SUBSTANCEMOLES OF A SINGLE SUBSTANCE
1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles = mass = 4g = 0.125 mol molar mass 32g mol -1
MOLES OF A SINGLE SUBSTANCEMOLES OF A SINGLE SUBSTANCE
MASS
MOLES x MOLAR MASS
1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles = mass = 4g = 0.125 mol molar mass 32g mol -1
MOLES OF A SINGLE SUBSTANCEMOLES OF A SINGLE SUBSTANCE
1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles = mass = 4g = 0.125 mol molar mass 32g mol -1
2. What is the mass of 0.25 mol of Na2CO3 ?
MOLES OF A SINGLE SUBSTANCEMOLES OF A SINGLE SUBSTANCE
1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles = mass = 4g = 0.125 mol molar mass 32g mol -1
2. What is the mass of 0.25 mol of Na2CO3 ?
Relative Molecular Mass of Na2CO3 = (2x23) + 12 + (3x16) = 106
Molar mass of Na2CO3 = 106g mol-1
MOLES OF A SINGLE SUBSTANCEMOLES OF A SINGLE SUBSTANCE
1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles = mass = 4g = 0.125 mol molar mass 32g mol -1
2. What is the mass of 0.25 mol of Na2CO3 ?
Relative Molecular Mass of Na2CO3 = (2x23) + 12 + (3x16) = 106
Molar mass of Na2CO3 = 106g mol-1
mass = moles x molar mass = 0.25 x 106 = 26.5g
MOLES OF A SINGLE SUBSTANCEMOLES OF A SINGLE SUBSTANCE
MASS
MOLES x MOLAR MASS
1. Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles = mass = 4g = 0.125 mol molar mass 32g mol -1
2. What is the mass of 0.25 mol of Na2CO3 ?
Relative Molecular Mass of Na2CO3 = (2x23) + 12 + (3x16) = 106
Molar mass of Na2CO3 = 106g mol-1
mass = moles x molar mass = 0.25 x 106 = 26.5g
MOLES OF A SINGLE SUBSTANCEMOLES OF A SINGLE SUBSTANCE
CaCO3 + 2HCl ———> CaCl2 + CO2 + H2O
1. What is the relative formula mass of CaCO3? 40 + 12 + (3 x 16) = 100
2. What is the mass of 1 mole of CaCO3 100 g
3. How many moles of HCl react with 1 mole of CaCO3? 2 moles
4. What is the relative formula mass of HCl? 35.5 + 1 = 36.5
5. What is the mass of 1 mole of HCl? 36.5 g
6. What mass of HCl will react with 1 mole of CaCO3 ? 2 x 36.5g = 73g
7. What mass of CO2 is produced ? moles of CO2 = moles of CaCO3
moles of CO2 = 0.001 moles
mass of CO2 = 0.001 x 44 = 0.044g
REACTING MASS CALCULATIONSREACTING MASS CALCULATIONS
EQUATIONS give you the ratio in which chemicals react and are formedneed to be balanced in order to do a calculation
CaCO3 + 2HCl ———> CaCl2 + CO2 + H2O
1. What is the relative molecular mass of CaCO3? 40 + 12 + (3 x 16) = 100
2. What is the mass of 1 mole of CaCO3? 100 g
3. What does 0.1M HCl mean? concentration is 0.1 mol dm-3
4. How many moles of HCl are in 20cm3 of 0.1M HCl? 0.1 x 20 = 0.002 moles 1000
5. How many moles of CaCO3 will react ? ½ x 0.002 = 0.001 moles
6. What is the mass of 0.001 moles of CaCO3? mass = moles x molar mass
= 0.001 x 100 = 0.1 g
7. What mass of CO2 is produced ? moles of CO2 = moles of CaCO3
moles of CO2 = 0.001 moles mass of CO2 = 0.001 x 44 = 0.044g
REACTING MASS CALCULATIONSREACTING MASS CALCULATIONS
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
UNITS concentration mol dm-3
volume dm3
BUT IF... concentration mol dm-3
volume cm3
THE MOLETHE MOLE
MOLES
CONC x VOLUME
COVER UP THE VALUE YOU WANT AND THE
METHOD OF CALCULATION IS
REVEALED
MOLES = CONCENTRATION x VOLUME MOLES = CONCENTRATION x VOLUME
MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)
MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3) 1000
MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3) 1000
250cm3
25cm3
250cm3
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
concentration of solution in the graduated flask = 0.100 mol dm-3
volume pipetted out into the conical flask = 25.00 cm3
THE MOLETHE MOLE
The original solution has a concentration of 0.100 mol dm-3
250cm3
25cm3
250cm3
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
concentration of solution in the graduated flask = 0.100 mol dm-3
volume pipetted out into the conical flask = 25.00 cm3
THE MOLETHE MOLE
The original solution has a concentration of 0.100 mol dm-3
This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solution
250cm3
25cm3
250cm3
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
concentration of solution in the graduated flask = 0.100 mol dm-3
volume pipetted out into the conical flask = 25.00 cm3
THE MOLETHE MOLE
The original solution has a concentration of 0.100 mol dm-3
This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solutionTake out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles
250cm3
25cm3
250cm3
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
concentration of solution in the graduated flask = 0.100 mol dm-3
volume pipetted out into the conical flask = 25.00 cm3
THE MOLETHE MOLE
The original solution has a concentration of 0.100 mol dm-3
This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solutionTake out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles
moles in 1dm3 (1000cm3) = 0.100moles in 1cm3 = 0.100/1000moles in 25cm3 = 25 x 0.100/1000 = 2.5 x 10-3 mol
250cm3
25cm3
250cm3
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
concentration of solution in the graduated flask = 0.100 mol dm-3
volume pipetted out into the conical flask = 25.00 cm3
THE MOLETHE MOLE
The original solution has a concentration of 0.100 mol dm-3
This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solutionTake out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles
moles in 1dm3 (1000cm3) = 0.100moles in 1cm3 = 0.100/1000moles in 25cm3 = 25 x 0.100/1000 = 2.5 x 10-3 mol
250cm3
25cm3
250cm3
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
concentration of solution in the graduated flask = 0.100 mol dm-3
volume pipetted out into the conical flask = 25.00 cm3
THE MOLETHE MOLE
1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH
MOLE OF SOLUTE IN A SOLUTIONMOLE OF SOLUTE IN A SOLUTION
MOLES = CONCENTRATION x VOLUMEMOLES = CONCENTRATION x VOLUME
1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH
moles = conc x volume in cm3
1000
= 2 mol dm-3 x 25cm3 = 0.05 moles 1000
MOLE OF SOLUTE IN A SOLUTIONMOLE OF SOLUTE IN A SOLUTION
MOLES = CONCENTRATION x VOLUMEMOLES = CONCENTRATION x VOLUME
1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH
moles = conc x volume in cm3
1000
= 2 mol dm-3 x 25cm3 = 0.05 moles 1000
2 What volume of 0.1M H2SO4 contains 0.002 moles ?
MOLE OF SOLUTE IN A SOLUTIONMOLE OF SOLUTE IN A SOLUTION
MOLES = CONCENTRATION x VOLUMEMOLES = CONCENTRATION x VOLUME
1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH
moles = conc x volume in cm3
1000
= 2 mol dm-3 x 25cm3 = 0.05 moles 1000
2 What volume of 0.1M H2SO4 contains 0.002 moles ?
volume = 1000 x moles (re-arrangement of above)
(in cm3) conc
= 1000 x 0.002 = 20 cm3
0.1 mol dm-3
MOLE OF SOLUTE IN A SOLUTIONMOLE OF SOLUTE IN A SOLUTION
MOLES = CONCENTRATION x VOLUMEMOLES = CONCENTRATION x VOLUME
Volumetric solutions are made by dissolving a known amount of solute in a solvent (usually water) and then adding enough solvent to get the correct volume of solution.
WRONGDissolve 1g of solute in 250cm3
of de-ionised water
SOLUTIONSSOLUTIONS
‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’
250cm3
1g
Volumetric solutions are made by dissolving a known amount of solute in a solvent (usually water) and then adding enough solvent to get the correct volume of solution.
WRONGDissolve 1g of solute in 250cm3
of de-ionised water
RIGHTDissolve 1g of solute in waterand then add enough water tomake 250cm3 of solution
SOLUTIONSSOLUTIONS
‘Dissolving a SOLUTE in a SOLVENT makes a SOLUTION’
250cm3
1g
250cm3
1g
WATER WATER
STANDARD SOLUTIONSTANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean The solution was transferred
beaker and dissolved in de-ionised water quantitatively to a 250 cm3
graduated flask and made upto the mark with de-ionised(or distilled) water.
STANDARD SOLUTIONSTANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean The solution was transferred
beaker and dissolved in de-ionised water quantitatively to a 250 cm3
graduated flask and made upto the mark with de-ionised(or distilled) water.
What is the concentration of the solution in mol dm-3 ?
STANDARD SOLUTIONSTANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean The solution was transferred
beaker and dissolved in de-ionised water quantitatively to a 250 cm3
graduated flask and made upto the mark with de-ionised(or distilled) water.
What is the concentration of the solution in mol dm-3 ?
mass of Na2CO3 in a 250cm3 solution = 4.240g
molar mass of Na2CO3 = 106g mol -1
no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol
STANDARD SOLUTIONSTANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean The solution was transferred
beaker and dissolved in de-ionised water quantitatively to a 250 cm3
graduated flask and made upto the mark with de-ionised(or distilled) water.
What is the concentration of the solution in mol dm-3 ?
mass of Na2CO3 in a 250cm3 solution = 4.240g
molar mass of Na2CO3 = 106g mol -1
no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol
Concentration is normally expressed as moles per dm3 of solutionTherefore, as it is in 250cm3, the value is scaled up by a factor of 4
STANDARD SOLUTIONSTANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean The solution was transferred
beaker and dissolved in de-ionised water quantitatively to a 250 cm3
graduated flask and made upto the mark with de-ionised(or distilled) water.
What is the concentration of the solution in mol dm-3 ?
mass of Na2CO3 in a 250cm3 solution = 4.240g
molar mass of Na2CO3 = 106g mol -1
no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol
Concentration is normally expressed as moles per dm3 of solutionTherefore, as it is in 250cm3, the value is scaled up by a factor of 4
no. of moles in 1000cm3 (1dm3) = 4 x 0.04 = 0.16 mol ANS. 0.16 mol dm-3
STANDARD SOLUTIONSTANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?
STANDARD SOLUTIONSTANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be? = 0.100 mol dm-3
How many moles will be in 1 dm3 ? = 0.100 molHow many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol
STANDARD SOLUTIONSTANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be? = 0.100 mol dm-3
How many moles will be in 1 dm3 ? = 0.100 molHow many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol
What is the formula of anhydrous sodium carbonate? = Na2CO3
What is the relative formula mass? = 106What is the molar mass? = 106g mol -1
STANDARD SOLUTIONSTANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be? = 0.100 mol dm-3
How many moles will be in 1 dm3 ? = 0.100 molHow many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol
What is the formula of anhydrous sodium carbonate? = Na2CO3
What is the relative formula mass? = 106What is the molar mass? = 106g mol -1
What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g
of Na2CO3 ? (mass = moles x molar mass)
STANDARD SOLUTIONSTANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be? = 0.100 mol dm-3
How many moles will be in 1 dm3 ? = 0.100 molHow many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol
What is the formula of anhydrous sodium carbonate? = Na2CO3
What is the relative formula mass? = 106What is the molar mass? = 106g mol -1
What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g
of Na2CO3 ? (mass = moles x molar mass)
ANS. The chemist will have to weigh out 2.650g, dissolve it in waterand then make the solution up to 250cm3 in a graduated flask.
STANDARD SOLUTIONSTANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be? = 0.100 mol dm-3
How many moles will be in 1 dm3 ? = 0.100 molHow many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol
What is the formula of anhydrous sodium carbonate? = Na2CO3
What is the relative formula mass? = 106What is the molar mass? = 106g mol -1
What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g
of Na2CO3 ? (mass = moles x molar mass)
ANS. The chemist will have to weigh out 2.650g, dissolve it in waterand then make the solution up to 250cm3 in a graduated flask.
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 isneutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 isneutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation NaOH + HCl ——> NaCl + H2O
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 isneutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation NaOH + HCl ——> NaCl + H2O
2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 isneutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation NaOH + HCl ——> NaCl + H2O
2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl
3. Calculate the number of moles of each substance HCl 0.100 x 25/1000 (i)
M is the concentration in mol dm-3 NaOH M x 20/1000 (ii)
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 isneutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation NaOH + HCl ——> NaCl + H2O
2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl
3. Calculate the number of moles of each substance HCl 0.100 x 25/1000 (i)
M is the concentration in mol dm-3 NaOH M x 20/1000 (ii)
4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl M x 20/1000 = 0.100 x 25/1000
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 isneutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation NaOH + HCl ——> NaCl + H2O
2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl
3. Calculate the number of moles of each substance HCl 0.100 x 25/1000 (i)
M is the concentration in mol dm-3 NaOH M x 20/1000 (ii)
4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl M x 20/1000 = 0.100 x 25/1000
5. Cancel the 1000’s M x 20 = 0.100 x 25
re-arrange the numbers to obtain M M = 0.100 x 25 20
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
Calculate the concentration of a solution of sodium hydroxide if 20cm3 isneutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3.
1. Write out a BALANCED equation NaOH + HCl ——> NaCl + H2O
2. Get a molar relationship between the reactants moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl
3. Calculate the number of moles of each substance HCl 0.100 x 25/1000 (i)
M is the concentration in mol dm-3 NaOH M x 20/1000 (ii)
4. Look at the molar relationship and insert (i) and (ii) moles of NaOH = moles of HCl M x 20/1000 = 0.100 x 25/1000
5. Cancel the 1000’s M x 20 = 0.100 x 25
re-arrange the numbers to obtain M M = 0.100 x 25 20
6. Calculate the concentration of the NaOH = 0.125 mol dm-3
Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake.
2NaOH + H2SO4 ——> Na2SO4 + 2H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
i.e moles of NaOH = 2 x moles of H2SO4
or moles of H2SO4 = moles of NaOH
2
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
REMEMBER... IT IS NOT A MATHEMATICAL EQUATION
2 moles of NaOH DO NOT EQUAL 1 mole of H2SO4
REMEMBER... IT IS NOT A MATHEMATICAL EQUATION
2 moles of NaOH DO NOT EQUAL 1 mole of H2SO4
More examples follow
Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake.
2NaOH + H2SO4 ——> Na2SO4 + 2H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
i.e moles of NaOH = 2 x moles of H2SO4
or moles of H2SO4 = moles of NaOH
2
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
2HCl + Na2CO3 ——> 2NaCl + CO2 + H2O
you need 2moles of HCl to react with every 1 mole of Na2CO3
i.e moles of HCl = 2 x moles of Na2CO3
or moles of Na2CO3 = moles of HCl
2
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake.
MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+
you need 5 moles of Fe2+ to react with every 1 mole of MnO4¯
i.e moles of Fe2+ = 5 x moles of MnO4¯
or moles of MnO4¯ = moles of Fe2+
5
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
2MnO4¯ + 5H2O2 + 6H+ ——> 2Mn2+ + 5O2 + 8H2O
you need 5 moles of H2O2 to react with every 2 moles of MnO4¯
i.e moles of H2O2 = 5 x moles of MnO4¯
2
or moles of MnO4¯ = 2 x moles of H2O2
5
Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake.
MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+
you need 5 moles of Fe2+ to react with every 1 mole of MnO4¯
i.e moles of Fe2+ = 5 x moles of MnO4¯
or moles of MnO4¯ = moles of Fe2+
5
Calculate the volume of sodium hydroxide (concentration 0.100 mol dm-3)required to neutralise 20cm3 of sulphuric acid of concentration 0.120 mol dm-3.
2NaOH + H2SO4 ——> Na2SO4 + 2H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
therefore moles of NaOH = 2 x moles of H2SO4
moles of H2SO4 = 0.120 x 20/1000 (i)
moles of NaOH = 0.100 x V/1000 (ii)
where V is the volume of alkali in cm3
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
Calculate the volume of sodium hydroxide (concentration 0.100 mol dm-3)required to neutralise 20cm3 of sulphuric acid of concentration 0.120 mol dm-3.
2NaOH + H2SO4 ——> Na2SO4 + 2H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
therefore moles of NaOH = 2 x moles of H2SO4
moles of H2SO4 = 0.120 x 20/1000 (i)
moles of NaOH = 0.100 x V/1000 (ii)
where V is the volume of alkali in cm3
substitute numbers moles of NaOH = 2 x moles of H2SO4
0.100 x V/1000 = 2 x 0.120 x 20/1000
cancel the 1000’s 0.100 x V = 2 x 0.120 x 20
re-arrange Volume of NaOH (V) = 2 x 0.120 x 20 = 48.00 cm3
0.100
VOLUMETRIC CALCULATIONSVOLUMETRIC CALCULATIONS
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 22.4dm3 at stp
1. Calculate the volume occupied by 0.25 mols of carbon dioxide at stp
1 mol of carbon dioxide will occupy a volume of 22.4 dm3 at stp0.25 mol of carbon dioxide will occupy a volume of 22.4 x 0.25 dm3 at stp0.25 mol of carbon dioxide will occupy a volume of 5.6 dm3 at stp
MOLAR VOLUMEMOLAR VOLUME
stp = standard temperature and pressure (273K and 105 Pa)ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 22.4dm3 at stp
1. Calculate the volume occupied by 0.25 mols of carbon dioxide at stp
1 mol of carbon dioxide will occupy a volume of 22.4 dm3 at stp0.25 mol of carbon dioxide will occupy a volume of 22.4 x 0.25 dm3 at stp0.25 mol of carbon dioxide will occupy a volume of 5.6 dm3 at stp
2. Calculate the volume occupied by 0.08g of methane (CH4) at stp
Relative Molecular Mass of CH4 = 12 + (4x1) = 16
Molar Mass of CH4 = 16g mol-1
Moles = mass/molar mass 0.08g / 16g mol-1 = 0.005 mols
1 mol of methane will occupy a volume of 22.4 dm3 at stp0.005 mol of carbon dioxide will occupy a volume of 22.4 x 0.005 dm3 at stp0.005 mol of carbon dioxide will occupy a volume of 0.112 dm3 at stp
MOLAR VOLUMEMOLAR VOLUME
stp = standard temperature and pressure (273K and 105 Pa)ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure
THE MOLETHE MOLETHE ENDTHE END
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